427(1,3,5) 428(1,3) помогите пожалуйста!

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Хуqожнuк_zn · Answer 1 · 2018-05-14T09:12:03+0000

427. 1
$625^{0.75}*243^{-0.4}-8^{-\frac{2}{3} }*27^{ \frac{1}{3} }+289^{0.5}= \\ \\ = 625^{ \frac{3}{4} }*( \frac{1}{243}) ^{\frac{2}{5} }-( \frac{1}{8} )^{\frac{2}{3} }*27^{ \frac{1}{3} }+289^{ \frac{1}{2} }= \\ \\ = 125* \frac{1}{9} - \frac{1}{4} *3+17= \frac{125}{9}- \frac{3}{4} +17=\frac{500}{36} - \frac{27}{36} +\frac{612}{36}= \frac{1085}{36} = 30 \frac{5}{36}$

427. 3
$((0.16)^{-5}*((6.25)^{-3})^2):((0.4)^{-2}*(2.5)^{-4})= \\ \\ =((0.4)^2 )^{-5}*((2.5)^2)^{-6}:((0.4)^{-2}*(2.5)^{-4})= \frac{(0.4)^{-10}*(2.5)^{-12}}{(0.4)^{-2}*(2.5)^{-4}} = \\ \\ =\frac{(0.4)^{2}*(2.5)^{4}}{(0.4)^{10}*(2.5)^{12}} = \frac{1}{(0.4)^8*(2.5)^8}= \frac{1}{(0.4*2.5)^8} = \frac{1}{1}=1$

427. 5
$125^{log_5 \frac{1}{2} -log_{125}2}=125^{log_5 \frac{1}{2} - \frac{1}{3} log_{5}2}=125^{log_5(2)^{-1} - log_{5}(2)^ \frac{1}{3} }= \\ \\ =5^{3log_5 \frac{(2)^{-1}}{(2)^ \frac{1}{3}} }=5^{3log_5(2)^{- \frac{4}{3} }}= 5^{log_5(2)^{-4 }}= 2^{-4}= \frac{1}{16}$

428. 1
$\frac{4*(\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2} } )^{-1}+3(\frac{\sqrt{2}+\sqrt{3} }{4 \sqrt{3}})^{-1}}{(6^{-1}+(\sqrt{6} )^{-1})^{-1}+(1+ (\sqrt{6}) ^{-1})^{-1}}=\frac{\frac{12 \sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{12\sqrt{3} }{\sqrt{2}+\sqrt{3}}}{(\frac{1}{6}+\frac{1}{\sqrt{6}})^{-1}+(1+\frac{1}{\sqrt{6} } )^{-1} } =$

$=\frac{\frac{12 \sqrt{2}+12 \sqrt{3} }{ \sqrt{2} +\sqrt{3} } }{( \frac{ \sqrt{6}+1 }{6})^{-1}+(\frac{ \sqrt{6}+1}{ \sqrt{6} } )^{-1}} = \frac{ \frac{12( \sqrt{2} + \sqrt{3} )}{ \sqrt{2} + \sqrt{3} } }{{ \frac{6 }{\sqrt{6}+1 }}+\frac{\sqrt{6} }{ \sqrt{6}+1}} =\frac{ \frac{12( \sqrt{2} + \sqrt{3} )}{ \sqrt{2} + \sqrt{3} } }{{ \frac{ \sqrt{6}( \sqrt{6} +1) }{\sqrt{6}+1 }}}= \frac{12}{ \sqrt{6} } = \sqrt{24} =2 \sqrt{6}$

428. 3
$(\frac{3}{\sqrt[3]{64}-\sqrt[3]{25} }+\frac{\sqrt[3]{40}}{\sqrt[3]{8}+\sqrt[3]{5} }- \frac{10}{\sqrt[3]{25} })^{-1}*(13-4\sqrt[3]{5}-2\sqrt[3]{25})+\sqrt[3]{25} =\\\\=( \frac{3}{ 4 -\sqrt[3]{25}}+\frac{\sqrt[3]{8*5}}{2+\sqrt[3]{5} }-\frac{10}{\sqrt[3]{25} })^{-1}*(13-4\sqrt[3]{5}-2\sqrt[3]{25})+\sqrt[3]{25} =\\\\=({\frac{3}{(2-\sqrt[3]{5})*(2+\sqrt[3]{5})}+ \frac{2\sqrt[3]{5} }{2+\sqrt[3]{5} }-\frac{10}{\sqrt[3]{25}})^{-1}*(13-4\sqrt[3]{5}-2\sqrt[3]{25} )+\sqrt[3]{25}=$

$=( \frac{3 \sqrt[3]{25}+2 \sqrt[3]{5} \sqrt[3]{25}(2- \sqrt[3]{5} ) -10(2- \sqrt[3]{5} )(2+ \sqrt[3]{5} ) }{ \sqrt[3]{25} (2- \sqrt[3]{5} )(2+ \sqrt[3]{5} )} )^{-1}* \\ \\ *(13-4\sqrt[3]{5}-2\sqrt[3]{25} )+\sqrt[3]{25} =\frac{\sqrt[3]{25} (2- \sqrt[3]{5} )(2+ \sqrt[3]{5} )}{3 \sqrt[3]{25}+2*5(2- \sqrt[3]{5} ) -10(4- \sqrt[3]{25} )} * \\ \\ *(13-4\sqrt[3]{5}-2\sqrt[3]{25} )+\sqrt[3]{25} = \frac{\sqrt[3]{25} (2- \sqrt[3]{5} )(2+ \sqrt[3]{5} )}{3 \sqrt[3]{25}+20- 10\sqrt[3]{5} -40+10\sqrt[3]{25} }*$

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